Partitions of convex planar regions into zonogons A zonogon is a centrally symmetric convex polygon.

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*Are there convex non-zonogons that can be partitioned into a finite number of (convex) zonogons?


*Same as 1 with the pieces allowed to be nonconvex but centrally symmetric polygons.
 A: I extend on the answer of Yaakov Baruch.
Choose a generic direction as being "down". We can then speak of bottom edges and top edges of the polygon and its tiles.
Suppose that the polygon $P$ is partitioned into convex tiles, so that there exists a perfect matching between the tiles that are not centrally symmetric and matched tiles are related by a central point reflection (see the figure below).
Assume further that the partition is edge-to-edge, that is, no vertex of one tile meets an edge of another tile (Yaakov explained why this is no restriction).
Let $e_0$ be a top edge.
Note that every edge is the top-edge of at most one tile.
We construct a sequence of edges $e_1,...,e_m$ of edges as follows:

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*if $e_i$ is the top edge of a centrally symmetric tile $T$, then let $e_{i+1}$ be the "opposite" bottom edge of $T$.

*if $e_i$ is the top edge of not-centrally symmetric tile $T$, then let $e_{i+1}$ be the corresponding edge in the matched tile to $T$. $e_{i+1}$ is then a bottom edge of the matched tile.

*if $e_i$ is not the top edge of any tile, then $e_i$ is a bottom edge of $P$.

Note that all edges in this sequence are of the same length and have the same orientation, and so each top edge of $P$ corresponds to an equally long and equally oriented bottom edge. Since $P$ is convex, the order of these edges at the bottom must be inverse to the order at the top. This shows that $P$ must be centrally symmetric.

Note that this formulation includes the case of non-convex tiles.
Every centrally symmetric tile can be decomposed into a set of convex centrally symmetric tiles and a set of pairs of matched convex tiles that are not necessarily centrally symmetric:

A: Any centrally symmetric convex polygon can be sudbivided into rhombuses, so assume only rhombuses are used. Now split them further (with lines parallel to the edges) so that no vertex of a rhombus falls in the middle of the edge of an adjacent rhombus. (This is clearly a finite process.) Then each bottom segment (according to a fixed arbitrary choice of vertical direction) into which the overall perimeter is now divided is matched with one and only one parallel top segment of the same length on the opposite side, by following a unique path edge-to-opposite-edge one rhombus at a time. By convexity, all parallel segments (of different length) are together on the both bottom and top sides (but possibly in different orders). Therefore the total length for each direction is the same on the two opposite sides.
UPDATE. The proof above of 1. can carry on to 2. as well. In fact there is no need for the components to be rhombi or to be convex, and no need to cut them. All that is needed instead is to label the segments into which an edge is split by a vertex, and to replicate that splitting and labeling in the opposite edge (notice: in a centrally symmetric way this time!). And so on, propagate all vertex points inside edges through all opposite parallel edges. When that is done, in a finite number of steps, once again thanks to the overall convexity, there will be paths creating a one-to-one (non order-preserving) correspondence between top and bottom segments with same direction and length, proving the overall central symmetry.

